Internal energy

General physics

🎯 Learning Objectives

16.1.1 Understand that internal energy is determined by the state of the system and that it can be expressed as the sum of a random distribution of kinetic and potential energies associated with the molecules of a system.
16.1.2 Relate a rise in temperature of an object to an increase in its internal energy (for ideal gases, directly proportional to temperature).
16.2.1 Recall and use W = p∆V for the work done when the volume of a gas changes at constant pressure.
16.2.1 Understand the difference between work done by the gas (expansion) and work done on the gas (compression).

🗣️ Language Objectives

Define internal energy, kinetic energy, potential energy of molecules, and work done by/on a gas using appropriate scientific vocabulary.
Explain how internal energy relates to the state of a system and its temperature.
Describe the calculation of work done during volume changes at constant pressure using W = p∆V.
Distinguish between positive and negative work in the context of gas expansion and compression.

🔑 Key Terms / Негізгі терминдер / Ключевые термины

English Term	Русский перевод	Қазақша аудармасы
Internal Energy (U)	Внутренняя энергия (U)	Ішкі энергия (U)
System	Система	Жүйе
State of a system	Состояние системы	Жүйенің күйі
Molecule	Молекула	Молекула
Kinetic Energy (of molecules)	Кинетическая энергия (молекул)	Кинетикалық энергия (молекулалардың)
Potential Energy (of molecules)	Потенциальная энергия (молекул)	Потенциалдық энергия (молекулалардың)
Random Distribution	Случайное распределение	Кездейсоқ таралуы
Work (W)	Работа (W)	Жұмыс (W)
Work done by the gas	Работа, совершаемая газом	Газдың атқарған жұмысы
Work done on the gas	Работа, совершаемая над газом	Газға түсірілген жұмыс
Pressure (p)	Давление (p)	Қысым (p)
Volume (V)	Объем (V)	Көлем (V)
Change in Volume (∆V)	Изменение объема (∆V)	Көлемнің өзгеруі (∆V)
Expansion	Расширение	Ұлғаю
Compression	Сжатие	Сығылу
Ideal Gas	Идеальный газ	Идеал газ

🃏 Flashcards for Practice

Review the key terms using flashcards. You can find relevant sets on Quizlet or create your own:

Search for «Internal Energy & Work Done by Gas» flashcards on Quizlet

Focus on understanding the definition and application of each term.

📖 Glossary / Глоссарий

Internal Energy (U): The sum of the random distribution of kinetic and potential energies of the molecules in a system. It is a state function, meaning it depends only on the current state of the system (e.g., its temperature, pressure, and volume), not on how the system reached that state.
Translation (Қазақша)
Жүйедегі молекулалардың кинетикалық және потенциалдық энергияларының кездейсоқ таралуының қосындысы. Бұл күй функциясы болып табылады, яғни ол жүйенің сол күйге қалай жеткеніне емес, тек жүйенің ағымдағы күйіне (мысалы, оның температурасы, қысымы және көлемі) байланысты.
Kinetic Energy of Molecules: Energy associated with the motion (translational, rotational, vibrational) of molecules. This is directly related to the temperature of the system.
Translation (Қазақша)
Молекулалардың қозғалысымен (ілгерілемелі, айналмалы, тербелмелі) байланысты энергия. Бұл жүйенің температурасымен тікелей байланысты.
Potential Energy of Molecules: Energy associated with the intermolecular forces and the bonds within molecules. For an ideal gas, intermolecular forces are assumed to be negligible, so the potential energy component is often considered zero or constant.
Translation (Қазақша)
Молекулааралық күштермен және молекулалар ішіндегі байланыстармен байланысты энергия. Идеал газ үшін молекулааралық күштер ескерілмейтіндей аз деп есептеледі, сондықтан потенциалдық энергия құраушысы көбінесе нөлге тең немесе тұрақты деп қарастырылады.
Work done BY a gas (W = p∆V at constant pressure): When a gas expands, it pushes against its surroundings (e.g., a piston) and does work on the surroundings. If the expansion occurs at constant pressure ‘p’ and the volume changes by ‘∆V’, the work done by the gas is W = p∆V. This work is positive if ∆V is positive (expansion).
Translation (Қазақша)
Газ ұлғайған кезде, ол қоршаған ортаға (мысалы, поршеньге) қысым түсіріп, қоршаған ортаға жұмыс атқарады. Егер ұлғаю ‘p’ тұрақты қысымда жүрсе және көлем ‘∆V’-ға өзгерсе, газдың атқарған жұмысы W = p∆V болады. Егер ∆V оң болса (ұлғаю), бұл жұмыс оң болады.
Work done ON a gas: When a gas is compressed, work is done on it by the surroundings. This work is positive (energy is added to the gas). If using the W = p∆V convention for work done *by* the gas, then during compression ∆V is negative, so W (work done by gas) is negative. The work done *on* the gas is then -W, which would be positive.
Translation (Қазақша)
Газ сығылған кезде, оған қоршаған орта жұмыс атқарады. Бұл жұмыс оң болады (газға энергия қосылады). Егер газдың атқарған жұмысы үшін W = p∆V конвенциясын қолдансақ, онда сығылу кезінде ∆V теріс болады, сондықтан W (газдың атқарған жұмысы) теріс болады. Сонда газға түсірілген жұмыс -W болады, ол оң мәнге ие болады.

🔬 Theory: Internal Energy & Work Done by/on a Gas / Теория: Ішкі энергия және Газдың атқарған/Газға түсірілген жұмыс

1. Internal Energy (U)

The internal energy of a system is a crucial concept in thermodynamics. It represents the total energy stored within the system due to its microscopic components – its molecules or atoms. Specifically, internal energy (U) is defined as:

U = Sum of random kinetic energies + Sum of random potential energies of the molecules

Let’s break this down:

Random Kinetic Energy: Molecules in any substance are in constant, random motion. This motion can be translational (moving from one place to another), rotational (spinning), and vibrational (atoms within molecules oscillating). The sum of these kinetic energies contributes to the internal energy. The average kinetic energy of the molecules is directly related to the temperature of the system.
Random Potential Energy: This arises from the intermolecular forces (forces between molecules) and, to a lesser extent, intramolecular forces (forces within molecules, i.e., chemical bonds, though these usually don’t change in simple thermodynamic processes). In solids and liquids, these forces are significant. In an ideal gas, intermolecular forces are assumed to be negligible, so the potential energy component due to these forces is taken as zero. Therefore, for an ideal gas, the internal energy is solely dependent on the kinetic energy of its molecules, and thus solely on its temperature.

Internal energy is determined by the state of the system (defined by variables like temperature, pressure, volume, and number of moles). It does not depend on the path taken to reach that state.

Relating Temperature Rise to Internal Energy Increase:

When the temperature of an object rises, the average kinetic energy of its molecules increases. Since internal energy includes this kinetic energy component, a rise in temperature generally leads to an increase in its internal energy.

For an ideal monatomic gas, the internal energy is directly proportional to its absolute temperature: U = (3/2)nRT, where n is the number of moles, R is the ideal gas constant, and T is the absolute temperature in Kelvin. For other substances, the relationship can be more complex but the general trend holds: higher temperature means higher internal energy.

2. Work Done When the Volume of a Gas Changes (W = p∆V)

When a gas expands or is compressed, work is done. Consider a gas enclosed in a cylinder with a movable piston.

Work Done BY the Gas (Expansion): If the gas expands, it pushes the piston outwards, exerting a force over a distance. Thus, the gas does work on its surroundings. If this expansion occurs at a constant pressure ‘p’, and the volume of the gas increases by ∆V (where ∆V = V_final — V_initial, and is positive for expansion), the work done by the gas is given by:
W = p∆V
In this case, W is positive, indicating energy is transferred from the gas to the surroundings.
Work Done ON the Gas (Compression): If an external force pushes the piston inwards, compressing the gas, work is done on the gas by the surroundings. The volume decreases, so ∆V is negative. Using the same formula W = p∆V, the work done by the gas will be negative (W < 0).
The work done on the gas is the negative of the work done by the gas. So, if W_{by gas} = p∆V (and ∆V is negative for compression), then W_{on gas} = — (p∆V). Since ∆V is negative, -p∆V becomes positive. This positive value signifies that energy is transferred from the surroundings to the gas.

Understanding the Sign Convention:

It’s important to be clear about the sign convention for work. In many physics contexts (especially when applying the First Law of Thermodynamics as ΔU = Q — W):

W > 0 means work is done BY the system (gas expands, energy leaves the system as work).
W < 0 means work is done ON the system (gas is compressed, energy enters the system as work).

If W = p∆V is defined as work done BY the gas:

Expansion: ∆V > 0 => W = p∆V > 0 (work done by gas is positive).
Compression: ∆V W = p∆V < 0 (work done by gas is negative). This means the work done ON the gas is -W, which is positive.

The formula W = p∆V is specifically for processes occurring at constant pressure (isobaric processes). If the pressure changes during the volume change, the work done is found by calculating the area under the pressure-volume (p-V) graph.

Theory Translation (Қазақша)

1. Ішкі энергия (U)

Жүйенің ішкі энергиясы – термодинамикадағы маңызды ұғым. Ол жүйенің микроскопиялық компоненттері – оның молекулалары немесе атомдары есебінен жүйе ішінде сақталған жалпы энергияны білдіреді. Нақтырақ айтқанда, ішкі энергия (U) былай анықталады:

U = Молекулалардың кездейсоқ кинетикалық энергияларының қосындысы + Молекулалардың кездейсоқ потенциалдық энергияларының қосындысы

Мұны талдап көрейік:

Кездейсоқ кинетикалық энергия: Кез келген заттағы молекулалар үнемі кездейсоқ қозғалыста болады. Бұл қозғалыс ілгерілемелі (бір жерден екінші жерге қозғалу), айналмалы (айналу) және тербелмелі (молекулалар ішіндегі атомдардың тербелуі) болуы мүмкін. Осы кинетикалық энергиялардың қосындысы ішкі энергияға үлес қосады. Молекулалардың орташа кинетикалық энергиясы жүйенің температурасымен тікелей байланысты.
Кездейсоқ потенциалдық энергия: Бұл молекулааралық күштерден (молекулалар арасындағы күштер) және аз дәрежеде молекулаішілік күштерден (молекулалар ішіндегі күштер, яғни химиялық байланыстар, бірақ бұлар қарапайым термодинамикалық процестерде әдетте өзгермейді) туындайды. Қатты денелер мен сұйықтықтарда бұл күштер маңызды. Идеал газда молекулааралық күштер ескерілмейтіндей аз деп есептеледі, сондықтан осы күштерге байланысты потенциалдық энергия құраушысы нөлге тең деп алынады. Сондықтан, идеал газ үшін ішкі энергия тек оның молекулаларының кинетикалық энергиясына, демек, тек оның температурасына байланысты болады.

Ішкі энергия жүйенің күйімен анықталады (температура, қысым, көлем және моль саны сияқты айнымалылармен анықталады). Ол сол күйге жету жолына байланысты емес.

Температураның көтерілуін ішкі энергияның артуымен байланыстыру:

Нысанның температурасы көтерілгенде, оның молекулаларының орташа кинетикалық энергиясы артады. Ішкі энергия осы кинетикалық энергия құраушысын қамтитындықтан, температураның көтерілуі әдетте оның ішкі энергиясының артуына әкеледі.

Идеал бір атомды газ үшін ішкі энергия оның абсолютті температурасына тура пропорционал: U = (3/2)nRT, мұндағы n – моль саны, R – идеал газ тұрақтысы, ал T – Кельвиндегі абсолютті температура. Басқа заттар үшін бұл байланыс күрделірек болуы мүмкін, бірақ жалпы үрдіс сақталады: температура жоғары болса, ішкі энергия да жоғары болады.

2. Газ көлемі өзгергенде атқарылатын жұмыс (W = p∆V)

Газ ұлғайғанда немесе сығылғанда жұмыс атқарылады. Қозғалмалы поршені бар цилиндрге қамалған газды қарастырайық.

Газдың атқарған жұмысы (Ұлғаю): Егер газ ұлғайса, ол поршеньді сыртқа қарай итеріп, белгілі бір қашықтықта күш түсіреді. Осылайша, газ қоршаған ортаға жұмыс атқарады. Егер бұл ұлғаю ‘p’ тұрақты қысымда жүрсе және газдың көлемі ∆V-ға артса (мұндағы ∆V = V_соңғы — V_{бастапқы}, және ұлғаю үшін оң мәнге ие), газдың атқарған жұмысы былай беріледі:
W = p∆V
Бұл жағдайда W оң мәнге ие, бұл энергияның газдан қоршаған ортаға тасымалданғанын білдіреді.
Газға түсірілген жұмыс (Сығылу): Егер сыртқы күш поршеньді ішке қарай итеріп, газды сықса, газға қоршаған орта жұмыс атқарады. Көлем азаяды, сондықтан ∆V теріс мәнге ие. W = p∆V формуласын қолдана отырып, газдың атқарған жұмысы теріс болады (W < 0).
Газға түсірілген жұмыс – газдың атқарған жұмысының теріс таңбамен алынғаны. Сонымен, егер W_{газ атқарған} = p∆V (және сығылу үшін ∆V теріс) болса, онда W_{газға түсірілген} = — (p∆V). ∆V теріс болғандықтан, -p∆V оң мәнге ие болады. Бұл оң мән энергияның қоршаған ортадан газға тасымалданғанын білдіреді.

Таңбалар ережесін түсіну:

Жұмыс үшін таңбалар ережесін нақты түсіну маңызды. Көптеген физика контекстерінде (әсіресе Термодинамиканың Бірінші Заңын ΔU = Q — W түрінде қолданғанда):

W > 0 жүйенің АТҚАРҒАН жұмысын білдіреді (газ ұлғаяды, энергия жүйеден жұмыс ретінде кетеді).
W < 0 жүйеге ТҮСІРІЛГЕН жұмысты білдіреді (газ сығылады, энергия жүйеге жұмыс ретінде кіреді).

Егер W = p∆V газдың АТҚАРҒАН жұмысы ретінде анықталса:

Ұлғаю: ∆V > 0 => W = p∆V > 0 (газдың атқарған жұмысы оң).
Сығылу: ∆V W = p∆V < 0 (газдың атқарған жұмысы теріс). Бұл газға ТҮСІРІЛГЕН жұмыстың -W екенін, яғни оң екенін білдіреді.

W = p∆V формуласы тұрақты қысымда (изобаралық процестер) жүретін процестер үшін арнайы. Егер көлем өзгерген кезде қысым өзгерсе, атқарылған жұмыс қысым-көлем (p-V) графигінің астындағы ауданды есептеу арқылы табылады.

Questions on Theory:

Easy: What are the two main forms of energy that contribute to the internal energy of a system?
Answer
The two main forms are the random kinetic energy of the molecules and the random potential energy of the molecules (due to intermolecular forces).
Medium: For an ideal gas, how does its internal energy change if its temperature increases? Why is the potential energy component often ignored for an ideal gas?
Answer
For an ideal gas, its internal energy increases if its temperature increases. This is because the internal energy of an ideal gas is solely dependent on the kinetic energy of its molecules, and average kinetic energy is directly proportional to absolute temperature. The potential energy component due to intermolecular forces is ignored because, by definition, the intermolecular forces in an ideal gas are assumed to be negligible.
Medium: A gas in a cylinder expands at a constant pressure of 200 kPa, and its volume increases by 0.05 m³. Calculate the work done by the gas. Is work done on the gas or by the gas?
Answer
Given: p = 200 kPa = 200 × 10³ Pa, ∆V = 0.05 m³.
Work done by the gas, W = p∆V
W = (200 × 10³ Pa) × (0.05 m³)
W = 10000 J or 10 kJ.
Since the gas expanded (∆V is positive), work is done by the gas.
Hard (Critical Thinking): If a gas is compressed slowly (quasi-statically) at a constant temperature (isothermal compression), work is done on the gas. Since the temperature does not change, the internal energy of an ideal gas does not change. According to the first law of thermodynamics (ΔU = Q + W_{on_gas}, where W_{on_gas} is work done on the gas), what must be happening to the heat (Q) during this isothermal compression?
Answer
For an isothermal compression of an ideal gas, the temperature remains constant, so the change in internal energy (ΔU) is zero.
The first law of thermodynamics can be written as ΔU = Q + W_{on_gas}, where Q is the heat added to the system and W_{on_gas} is the work done on the system.
Since ΔU = 0, we have 0 = Q + W_{on_gas}.
During compression, work is done on the gas, so W_{on_gas} is positive.
Therefore, Q = -W_{on_gas}.
This means Q must be negative. A negative Q indicates that heat is being removed from the gas (or flows out of the gas) during the isothermal compression. The amount of heat removed is equal to the amount of work done on the gas to keep its temperature (and thus internal energy for an ideal gas) constant.

🧠 Exercises on Memorizing Terms

Activity 1: Define in Your Own Words

For each term below, write a short definition in your own words:

Internal Energy
Work done by a gas
Ideal Gas (in terms of internal energy)
p∆V

Example Answers for Activity 1

Internal Energy: The total microscopic energy stored in a substance, which includes the kinetic energy from the motion of its molecules and the potential energy from the forces between them.
Work done by a gas: The energy transferred when a gas expands and pushes against something, like a piston.
Ideal Gas (in terms of internal energy): A gas where the molecules have no intermolecular forces, so its internal energy is only due to their kinetic energy (and thus only depends on temperature).
p∆V: A formula to calculate the work done by a gas (or on a gas, depending on sign convention) when it changes volume by ∆V at a constant pressure p.

Activity 2: True or False?

The internal energy of a system depends on the path taken to reach its current state. (True/False)
When a gas expands, it does positive work on its surroundings. (True/False)
For an ideal gas, if the temperature remains constant, its internal energy must also remain constant. (True/False)
If work is done on a gas, its volume must have increased. (True/False)
The potential energy of molecules in an ideal gas is very significant. (True/False)

Answers for Activity 2

False (Internal energy is a state function, independent of path.)
True (Work done by the gas W = p∆V; for expansion ∆V > 0, so W > 0.)
True (For an ideal gas, U depends only on T.)
False (If work is done on a gas, it is compressed, so its volume decreases.)
False (For an ideal gas, intermolecular potential energy is assumed to be negligible/zero.)

📺 Watch & Learn: Video on Internal Energy & Work Done

This video explains internal energy and work done by a gas:

Further Viewing — Related Topics:

📝 Solved Examples / Есептерді шешу мысалдары

Example 1: Internal Energy Change of an Ideal Gas

An ideal monatomic gas is heated at constant volume, causing its temperature to rise from 27°C to 127°C.

(a) Convert the initial and final temperatures to Kelvin.

(b) How does the internal energy of the gas change (increase, decrease, or stay the same)? Explain your reasoning.

(c) If the gas were diatomic instead of monatomic, would the change in internal energy for the same temperature rise be smaller, larger, or the same? (Conceptual explanation needed).

Brief Solution (Example 1)Detailed Solution (Example 1)

(a) T_initial = 300 K, T_final = 400 K.

(b) Internal energy increases because temperature increases, and for an ideal gas, U is directly proportional to T.

(a) Convert temperatures to Kelvin:

Temperature in Kelvin (T_K) = Temperature in Celsius (T_°C) + 273.15 (often approximated as 273)

Initial Temperature (T_initial): 27°C + 273 = 300 K

Final Temperature (T_final): 127°C + 273 = 400 K

(Textual pronunciation: Initial temperature is twenty-seven degrees Celsius plus two hundred seventy-three, which equals three hundred Kelvin. Final temperature is one hundred twenty-seven degrees Celsius plus two hundred seventy-three, which equals four hundred Kelvin.)

(b) Change in internal energy:

The internal energy of an ideal gas depends solely on its temperature. Since the temperature of the gas rises (from 300 K to 400 K), the average kinetic energy of its molecules increases. Therefore, the internal energy of the gas increases.

(Textual pronunciation: The internal energy of the gas increases because its temperature increases.)

For the same temperature rise, the change in internal energy for a diatomic gas would be larger than for a monatomic gas (assuming the same number of moles). This is because diatomic molecules have more degrees of freedom (translational, rotational, and at higher temperatures, vibrational) to store energy compared to monatomic gases (which primarily have translational kinetic energy). Each degree of freedom contributes to the internal energy. So, to achieve the same temperature rise (same average translational KE per molecule), more total energy needs to be added to a diatomic gas to also energize its rotational (and possibly vibrational) modes.

(Textual pronunciation: The change in internal energy for a diatomic gas would be larger because diatomic molecules have more ways to store energy, such as rotation, in addition to translational motion.)

Example 2: Work Done During Isobaric Expansion

A gas is contained in a cylinder fitted with a frictionless piston. The gas slowly expands from an initial volume of 2.0 × 10^-3 m³ to a final volume of 5.0 × 10^-3 m³ while its pressure remains constant at 1.5 × 10⁵ Pa.

(a) Calculate the change in volume (∆V).

(b) Calculate the work done by the gas during this expansion.

(c) If the process were a compression from 5.0 × 10^-3 m³ to 2.0 × 10^-3 m³ at the same constant pressure, what would be the work done by the gas? What would be the work done on the gas?

Brief Solution (Example 2)Detailed Solution (Example 2)

(a) ∆V = 3.0 × 10^-3 m³

(b) Work done by gas = 450 J

Given: Initial volume V₁ = 2.0 × 10^-3 m³, Final volume V₂ = 5.0 × 10^-3 m³, Constant pressure p = 1.5 × 10⁵ Pa.

(a) Calculate the change in volume (∆V):

∆V = V_final — V_initial = V₂ — V₁

∆V = (5.0 × 10^-3 m³) — (2.0 × 10^-3 m³)

∆V = 3.0 × 10^-3 m³

(Textual pronunciation: Delta V equals final volume minus initial volume, which is five point zero times ten to the power of minus three cubic meters minus two point zero times ten to the power of minus three cubic meters, equals three point zero times ten to the power of minus three cubic meters.)

(b) Calculate the work done by the gas during this expansion:

Work done by the gas at constant pressure is W = p∆V.

W = (1.5 × 10⁵ Pa) × (3.0 × 10^-3 m³)

W = 4.5 × 10^(5-3) J

W = 4.5 × 10² J = 450 J

Since ∆V is positive (expansion), the work done by the gas is positive.

(Textual pronunciation: Work equals p times delta V. This is one point five times ten to the power of five Pascals multiplied by three point zero times ten to the power of minus three cubic meters, which equals four hundred fifty Joules.)

In this case, V_initial = 5.0 × 10^-3 m³ and V_final = 2.0 × 10^-3 m³.

∆V = V_final — V_initial = (2.0 × 10^-3 m³) — (5.0 × 10^-3 m³) = -3.0 × 10^-3 m³.

Work done by the gas: W_{by gas} = p∆V = (1.5 × 10⁵ Pa) × (-3.0 × 10^-3 m³) = -450 J.

A negative value for work done by the gas means that work was actually done on the gas.

Work done on the gas: W_{on gas} = — W_{by gas} = -(-450 J) = 450 J.

(Textual pronunciation: For compression, delta V is minus three point zero times ten to the power of minus three cubic meters. Work done by the gas is p times delta V, which is minus four hundred fifty Joules. Work done on the gas is the negative of this, which is positive four hundred fifty Joules.)

💻 Investigation Task: Gas Properties Simulation / Зерттеу тапсырмасы: Газ қасиеттері симуляциясы

Explore the relationship between pressure, volume, temperature, and work done using the PhET Interactive Simulation «Gas Properties».

Link to Simulation: PhET Gas Properties Simulation

Instructions:

Open the simulation. You can start with the «Ideal» tab.
Pump some gas particles into the container.
Observe the pressure, volume, and temperature readings. You can turn on «Width» under «Measurement Tools» to see the container width (related to volume).
You can change the volume by dragging the handle on the left side of the container. You can add or remove heat using the control at the bottom.

Tasks:

Task 1 (Constant Pressure Expansion):
Try to keep the pressure approximately constant (you might need to add heat as you expand or remove heat as you compress to do this perfectly, or select «Pressure» in the «Constant Parameter» section on the right if available and you manipulate volume/heat).
Slowly increase the volume of the container. What do you observe happening to the gas particles? If you were to calculate W = p∆V, would the work done by the gas be positive or negative?
Task 2 (Constant Pressure Compression):
Now, slowly decrease the volume of the container, again trying to keep pressure constant. What happens to the gas particles? Would the work done by the gas be positive or negative? What about the work done on the gas?
Task 3 (Relating Temperature and Internal Energy):
Keep the volume constant. Add heat to the container. What happens to the temperature of the gas? What does this imply about the internal energy of the gas? Observe the speed of the particles.

Brief Answers & Observations

Task 1 Answer (Constant Pressure Expansion):

As the volume increases, the gas particles spread out to fill the larger space. To keep pressure constant, if you are manually controlling, you might need to add heat, which would make the particles move faster to maintain the same rate of collisions with the walls over a larger area.
Since ∆V is positive (volume increases), the work done by the gas (W = p∆V) would be positive.

(Қазақша: Көлем артқанда, газ бөлшектері үлкенірек кеңістікті толтыру үшін таралады. Қысымды тұрақты ұстау үшін, егер қолмен басқарсаңыз, жылу қосу қажет болуы мүмкін, бұл бөлшектердің қабырғалармен соқтығысу жылдамдығын үлкенірек ауданда сақтау үшін тезірек қозғалуына әкеледі. ∆V оң болғандықтан (көлем артады), газдың атқарған жұмысы (W = p∆V) оң болады.)

Task 2 Answer (Constant Pressure Compression):

As the volume decreases, the gas particles are confined to a smaller space. To keep pressure constant, you might need to remove heat, which would slow down the particles to maintain the same rate of collisions.
Since ∆V is negative (volume decreases), the work done by the gas (W = p∆V) would be negative.
The work done on the gas would be positive (equal to -W_{by gas}).

(Қазақша: Көлем азайғанда, газ бөлшектері кішірек кеңістікке қамалады. Қысымды тұрақты ұстау үшін, жылуды алып тастау қажет болуы мүмкін, бұл соқтығысу жылдамдығын сақтау үшін бөлшектерді баяулатады. ∆V теріс болғандықтан (көлем азаяды), газдың атқарған жұмысы (W = p∆V) теріс болады. Газға түсірілген жұмыс оң болады (-W_{газ атқарған}-ға тең).)

Task 3 Answer (Temperature and Internal Energy):

When heat is added at constant volume, the temperature of the gas increases.
This implies that the internal energy of the gas has increased, as for an ideal gas, internal energy is directly related to temperature.
The particles will be observed to move faster on average.

(Қазақша: Тұрақты көлемде жылу қосылғанда, газдың температурасы артады. Бұл газдың ішкі энергиясының артқанын білдіреді, себебі идеал газ үшін ішкі энергия температурамен тікелей байланысты. Бөлшектердің орташа жылдамдығының артқаны байқалады.)

🤝 Pair/Group Work: P-V Diagram Challenge / Жұптық/Топтық жұмыс: P-V диаграммасы бойынша тапсырма

Work with a partner or in a small group.

Task: Sketch and Interpret P-V Diagrams

Consider the following processes for an ideal gas:

Isobaric Expansion: The gas expands from volume V₁ to V₂ at a constant pressure P₁.
Isobaric Compression: The gas is compressed from volume V₂ to V₁ at a constant pressure P₁.

For each process:

a) Sketch the process on a P-V diagram (Pressure on the y-axis, Volume on the x-axis). Clearly label the initial and final states and the direction of the process.

b) How would you determine the work done by the gas from this P-V diagram?

c) Is the work done by the gas positive, negative, or zero? Explain.

d) What happens to the internal energy of the gas if its temperature increases during the expansion? What if its temperature decreases during compression (assuming heat is removed)?

Discuss your sketches and interpretations within your group. You can use Formative or a shared whiteboard tool to sketch and collaborate if working remotely.

Guidance for P-V Diagram Challenge

1. Isobaric Expansion (V₁ to V₂ at P₁):

a) Sketch: A horizontal line on the P-V diagram at pressure P₁, moving from V₁ to V₂ (arrow pointing right).

b) Work done: The area under the horizontal line (which is a rectangle). Area = P₁ × (V₂ — V₁) = P₁∆V.

c) Positive: Since V₂ > V₁, ∆V is positive, so work done by the gas is positive.

d) If temperature increases during expansion, internal energy increases (ΔU > 0). This means heat must have been added (Q > 0) such that Q > W_{by gas}.

2. Isobaric Compression (V₂ to V₁ at P₁):

a) Sketch: A horizontal line on the P-V diagram at pressure P₁, moving from V₂ to V₁ (arrow pointing left).

b) Work done: The area under the horizontal line. Area = P₁ × (V₁ — V₂). Since V₁ < V₂, this ∆V is negative.

c) Negative: Work done by the gas = P₁∆V is negative. This means work is done on the gas, and that work is positive.

d) If temperature decreases during compression, internal energy decreases (ΔU < 0). Work is done on the gas (W_{on gas} > 0). For ΔU to be negative, significant heat must be removed (Q < W_{on gas}.

✍️ Individual Work: Structured Questions / Жеке жұмыс: Құрылымдық сұрақтар

Answer the following questions to test your understanding and ability to apply the concepts of internal energy and work done. Show your working where necessary.

An ideal gas undergoes a process where its internal energy increases by 500 J.
(a) If 800 J of heat is supplied to the gas, calculate the work done by the gas. (Use the first law of thermodynamics: ΔU = Q — W, where W is work done by the gas).
(b) If, in a different process, the internal energy of the same ideal gas increases by 500 J while the gas does 200 J of work on its surroundings, how much heat was supplied to the gas?
A fixed mass of an ideal gas is heated at constant volume. Explain, in terms of molecular motion and energy:
(i) Why its temperature increases.
(ii) Why its internal energy increases.
(iii) Why no work is done by the gas.
A cylinder contains 0.020 m³ of a gas at a pressure of 2.5 × 10⁵ Pa. The gas is compressed at this constant pressure until its volume is 0.012 m³.
(a) Calculate the work done by the gas.
(b) Calculate the work done on the gas.
(c) If the internal energy of the gas decreases by 100 J during this compression, determine the magnitude and direction of heat flow. (Use ΔU = Q — W_{by_gas}).
Distinguish clearly between the internal energy of a gas and the heat supplied to a gas. Under what circumstances can heat be supplied to an ideal gas without changing its internal energy?
The internal energy of a real gas depends on both its temperature and its volume, whereas for an ideal gas, it depends only on temperature. Explain why the volume dependence arises for a real gas but not for an ideal gas, considering molecular potential energy.

(Note: These questions are designed for analysis and synthesis. Focus on clear explanations and accurate calculations.)

Answers to Selected Individual Work Questions (Guidance)

Guidance for Q1: (ΔU = Q — W_{by_gas})

(a) ΔU = +500 J, Q = +800 J. 500 = 800 — W_{by_gas} => W_{by_gas} = 800 — 500 = 300 J.

(b) ΔU = +500 J, W_{by_gas} = +200 J. 500 = Q — 200 => Q = 500 + 200 = 700 J.

Guidance for Q2:

(i) Heat supplied increases molecular KE => temperature increases.

(ii) Increased molecular KE => increased internal energy (as U for ideal gas is sum of KE).

(iii) Constant volume means ∆V = 0. Since W = p∆V, W = 0.

Guidance for Q3:

(a) p = 2.5 × 10⁵ Pa. V_initial = 0.020 m³, V_final = 0.012 m³.
∆V = V_final — V_initial = 0.012 — 0.020 = -0.008 m³.
W_{by_gas} = p∆V = (2.5 × 10⁵ Pa) × (-0.008 m³) = -2000 J.

(b) W_{on_gas} = -W_{by_gas} = -(-2000 J) = 2000 J.

(c) ΔU = -100 J (decreases). ΔU = Q — W_{by_gas}.
-100 J = Q — (-2000 J) => -100 J = Q + 2000 J => Q = -100 J — 2000 J = -2100 J.
Heat flows out of the gas (2100 J removed).

Guidance for Q4:

Internal energy is energy *stored* in the gas (sum of KE and PE of molecules). Heat is energy *transferred* due to temperature difference.
Heat can be supplied without changing U if the gas does an equal amount of work (isothermal expansion, ΔU=0, so Q=W_{by_gas}).

Guidance for Q5:

Real gas: Intermolecular forces exist, so molecular PE depends on average separation of molecules (i.e., volume). Changing volume changes PE, thus U.
Ideal gas: Intermolecular forces are negligible, so PE is negligible/zero. U only depends on KE, which depends on T.

📚 Further Resources & Reading / Қосымша ресурстар және оқу материалдары

Save My Exams — Internal Energy: Internal Energy Notes
Save My Exams — Work Done by a Gas: Work Done by a Gas Notes
PhysicsAndMathsTutor — First Law of Thermodynamics: A-Level CIE Physics — First Law of Thermodynamics (includes U and W)
OpenStax University Physics — Internal Energy: Chapter 3.3: Internal Energy and Heat
OpenStax University Physics — Work and PV Diagrams: Chapter 3.4: Work and PV Diagrams

🤔 Lesson Reflection / Сабақ бойынша рефлексия

Take a few moments to reflect on what you have learned in this lesson:

What is the key difference between the internal energy of an ideal gas and a real gas?
If a gas expands and its temperature drops, what can you say about the heat supplied to it relative to the work done by it? (Hint: Think about ΔU = Q — W).
Why is the formula W = p∆V only applicable for processes at constant pressure? How would you find work done if pressure is not constant?
On a scale of 1 (Not at all understood) to 5 (Very well understood), how would you rate your current understanding of the sign conventions for work done by a gas versus work done on a gas?
What real-world application or device can you think of where understanding the work done by or on a gas is crucial?